Embibe Experts Solutions for Chapter: Chemical Equilibrium & Chemical Kinetics, Exercise 2: Exercise 2

Steam reacts with iron at high temperature to give hydrogen gas and ${\mathrm{Fe}}_3{\mathrm{O}}_4 \left(\mathrm{s}\right).$ The correct expression for the equilibrium constant is

$($The reaction is:$3\mathrm{Fe}{ }_{\left(\mathrm{s}\right)}+4{\mathrm{H}}_2\mathrm{O} \left(\mathrm{steam}\right)\rightleftharpoons {\mathrm{Fe}}_3{\mathrm{O}}_{4 \left(\mathrm{s}\right)}+4{\mathrm{H}}_2 \left(\mathrm{g}\right)$$)$

The reaction of hydrogen and iodine monochloride is given as:

${\mathrm{H}}_{2 \left(\mathrm{g}\right)}+2{\mathrm{ICl}}_{ \left(\mathrm{g}\right)}⟶2{\mathrm{HCl}}_{ \left(\mathrm{g}\right)}+{\mathrm{I}}_{2 \left(\mathrm{g}\right)}$

The reaction is of a first-order, with respect to ${\mathrm{H}}_{2 \left(\mathrm{g}\right)}$ and ${\mathrm{ICl}}_{ \left(\mathrm{g}\right)},$ following mechanisms were proposed.

Mechanism $\mathrm{A}:$

${\mathrm{H}}_{2 \left(\mathrm{g}\right)}+2\mathrm{ICl}{ }_{\left(\mathrm{g}\right)}⟶2{\mathrm{HCl}}_{ \left(\mathrm{g}\right)}+{\mathrm{I}}_{2 \left(\mathrm{g}\right)}$

Mechanism $\mathrm{B}:$

${\mathrm{H}}_{2 \left(\mathrm{g}\right)}+{\mathrm{ICl}}_{ \left(\mathrm{g}\right)}⟶{\mathrm{HI}}_{ \left(\mathrm{g}\right)};$ slow

$\mathrm{HI}{ }_{\left(\mathrm{g}\right)}+{\mathrm{ICl}}_{ \left(\mathrm{g}\right)}⟶\mathrm{HCl}{ }_{\left(\mathrm{g}\right)}+{\mathrm{I}}_{2 \left(\mathrm{g}\right)};$ fast

Which of the above mechanism(s) can be consistent with the given information about the reaction?

;In the reversible reaction, $2 {\mathrm{NO}}_2\underset{{\mathrm{k}}_2}{\overset{{\mathrm{k}}_1}{\rightleftharpoons }}{\mathrm{N}}_2{\mathrm{O}}_4$ the rate of disappearance of ${\mathrm{NO}}_2$ is equal to:

The reaction,

$2 \mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons 3\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$

is begun with the concentrations of $\mathrm{A}$ and $\mathrm{B}$ both at an initial value of $1.00 \mathrm{M}.$ When equilibrium is reached, the concentration of $\mathrm{D}$ is measured and found to be $0.25 \mathrm{M}.$ The value for the equilibrium constant for this reaction is given by the expression:

The hypothetical reaction,

${\mathrm{A}}_2+{\mathrm{B}}_2⟶2\mathrm{AB};$

follows the following mechanism:

${\mathrm{A}}_2\stackrel{\mathrm{Fast}}{\rightleftharpoons }\mathrm{A}+\mathrm{A},$

$\mathrm{A}+{\mathrm{B}}_2\stackrel{\text{ Slow }}{⟶}\mathrm{AB}+\mathrm{B},$

The order of the overall reaction is:

For the reaction system,

$2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)⟶2{\mathrm{NO}}_2\left(\mathrm{g}\right)$ volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to ${\mathrm{O}}_2$ and second order with respect to $\mathrm{NO},$ the rate of reaction will:

Consider the following reaction,

The reaction is of first order in each diagram, with an equilibrium constant of ${10}^4.$ For the conversion of chair form to boat form ${\mathrm{e}}^{-\mathrm{Ea}/\mathrm{RT}}=4.35\times {10}^{-8}$ at $298 \mathrm{K}$ with pre-exponential factor of ${10}^{12} {\mathrm{s}}^{-1}.$ Apparent rate constant $\left({\mathrm{k}}_{\mathrm{A}}\right)$ at $298 \mathrm{K}$ is:

Acid hydrolysis of ester is a first order reaction and the rate constant is given by, $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{V}}_{\infty }-{\mathrm{V}}_{\circ }}{{\mathrm{V}}_{\infty }-{\mathrm{V}}_{\mathrm{t}}}$

where, ${\mathrm{V}}_0, {\mathrm{V}}_{\mathrm{t}}$ and ${\mathrm{V}}_{\infty }$ are the volume of standard $\mathrm{NaOH}$ required to neutralise an acid present. At any given time, if ester is $50\%$ neutralised at time $\mathrm{t}$, then